Worked examples — Transfer function — Laplace domain, poles and zeros

Before symbols: a quick reminder of the alphabet we use, so nothing appears un-earned.

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The scenario matrix

Every transfer function you meet falls into one of these cells. Each example below is tagged with the cell it covers.

#	Cell (case class)	What makes it distinct	Example
A	Two distinct real poles, both left	$\sigma <0$, $\omega =0$	Ex 1
B	Complex-conjugate poles, left	$\sigma <0$, $\omega \ne 0$ → ringing	Ex 2
C	Right-half-plane pole	some $\sigma >0$ → unstable	Ex 3
D	Pole on the imaginary axis (integrator / pure oscillator)	$\sigma =0$ → marginal	Ex 4
E	Repeated (degenerate) pole	double root, $t{e}^{pt}$ term	Ex 5
F	A zero present — does it change stability?	zero reweights, poles decide	Ex 6
G	Right-half-plane zero (non-minimum-phase)	"wrong-way" initial motion	Ex 7
H	Real-world word problem (rocket actuator)	build $G$ from an ODE, read personality	Ex 8
I	Limiting values (DC gain, high-freq roll-off)	$s\to 0$ and $s\to \infty$	Ex 9
J	Exam twist — a parameter hides the instability	find the threshold	Ex 10

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Ex 1 — Cell A: two distinct real poles (both stable)

Forecast: Both terms in the denominator are positive with positive coefficients — guess: two negative real roots → decaying, no wiggle.

1. Factor the denominator. Why this step? Poles are the roots of $D(s)=0$; factoring exposes them. $s^2+5s+6=(s+2)(s+3)$ Poles at $s=-2$ and $s=-3$.
2. Look for zeros. Why? Numerator is the constant $4$; $4=0$ has no solution, so no finite zeros.
3. Read the signs. Why? Real part of each pole is the decay rate. Both $\sigma =-2,-3<0$ → stable. Both $\omega =0$ → no oscillation.
4. Write the shape. Response $\propto c_1{e}^{-2t}+c_2{e}^{-3t}$. The slower pole $s=-2$ (closer to the imaginary axis) dominates; time constant $\tau =1/2=0.5$ s.

Verify: Steady-state gain (set $s=0$): $G(0)=4/6=0.667$. A step input of size $1$ settles to $0.667$ — finite, matching a stable system. ✓

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Ex 2 — Cell B: complex-conjugate poles (ringing)

Forecast: The discriminant looks like it will go negative → complex poles → the thing should ring while decaying.

1. Match the standard 2nd-order form $s^2+2\zeta {\omega }_{n}s+{\omega }_n^2$. Why? This is the standard 2nd-order model — reading $\zeta$ and ${\omega }_n$ off it tells us everything. ${\omega }_n^2=25\Rightarrow {\omega }_n=5$ rad/s. And $2\zeta {\omega }_n=4\Rightarrow \zeta =4/10=0.4$.
2. Solve the quadratic. Why? Poles are $s=-\zeta {\omega }_n\pm j{\omega }_n\sqrt{1-{\zeta }^2}$. $s=-0.4(5)\pm j5\sqrt{1-0.16}=-2\pm j5(0.9165)=-2\pm j4.583$
3. Read it. Real part $-2<0$ → decays. Imaginary part $\pm 4.583$ → rings at ${\omega }_d=4.583$ rad/s (the damped frequency). Since $0<\zeta <1$, this is underdamped: overshoot then settling.

Look at the red dots in the figure: they sit left of the vertical axis (stable) but off the horizontal axis (oscillating). Their distance from the origin is ${\omega }_n=5$; the angle from the vertical axis has cosine $\zeta$.

Verify: $|\text{pole}|=\sqrt{2^2+4.583^2}=\sqrt{4+21}=\sqrt{25}=5={\omega }_n$. ✓ And $\cos (\text{angle from real axis... measured from negative real axis})=2/5=0.4=\zeta$. ✓

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Ex 3 — Cell C: a right-half-plane pole (unstable)

Forecast: That $-2s$ (a negative middle coefficient) is a red flag — one root may sneak into the right half. Guess: unstable.

1. Factor. Why? Expose the poles. $s^2-2s-8=(s-4)(s+2)$ Poles at $s=+4$ and $s=-2$.
2. Check signs. Why? Any $\sigma >0$ means a growing $e^{pt}$. Pole $s=+4$ has $\sigma =+4>0$ → the term $e^{+4t}$ blows up.
3. Conclusion: one RHP pole ⇒ UNSTABLE, no matter how stable the other pole is.

Verify (sign test): For a 2nd-order polynomial $s^2+a_{1}s+a_0$, all roots are left-half only if both $a_1>0$ and $a_0>0$ (the Routh-Hurwitz shortcut). Here $a_1=-2<0$ → fails → unstable, agreeing with the factoring. ✓

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Ex 4 — Cell D: pole on the imaginary axis (marginal)

Forecast: No decay term anywhere → neither blows up nor settles → marginally stable (borderline).

1. (a) Integrator. Why look at $s$ alone? One pole at $s=0$: $\sigma =0$, $\omega =0$. Its time term is $e^{0\cdot t}=1$ — constant, neither growing nor decaying. A step input $1/s$ gives output $1/s^2$, which inverts to $t$ (a ramp that grows without bound). So an integrator is marginally stable: bounded inputs can produce unbounded output.
2. (b) Oscillator. Why? $s^2+9=0\Rightarrow s=\pm j3$. Real part $\sigma =0$, imaginary part $\pm 3$. Term $e^{j3t}$ oscillates forever at $3$ rad/s with no decay — a bell that never stops.
3. Conclusion: both live on the imaginary axis → marginally stable, the knife-edge between Cells A/B (left, stable) and Cell C (right, unstable).

Verify: For (b), poles at $\pm j3$ satisfy $s^2+9=0$: $(j3{)}^2+9=-9+9=0$. ✓ Real part exactly $0$. ✓

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Ex 5 — Cell E: a repeated (degenerate) pole

Forecast: Two poles stacked at the same spot $s=-2$. Still left-half → stable, but something changes in the shape.

1. Identify the double pole. Why? $(s+2{)}^2=0$ gives $s=-2$ twice — a repeated root.
2. Recall the inverse transform for a double pole. Why this and not a single-pole rule? A single pole gives ${\mathcal{L}}^{-1}\{1/(s+2)\}=e^{-2t}$, but a repeated pole gives an extra factor of $t$: ${\mathcal{L}}^{-1}\left\{\frac{1}{(s+2{)}^2}\right\}=t{e}^{-2t}$
3. Interpret. The $t$ multiplies, but $e^{-2t}$ decays faster than $t$ grows, so the product still → $0$. Stable, non-oscillatory, but with a slower "critically-damped-like" rise (this is exactly the $\zeta =1$ boundary).

Verify: $t{e}^{-2t}$ peaks where its derivative is zero: $\frac{d}{dt}(t{e}^{-2t})=e^{-2t}(1-2t)=0\Rightarrow t=0.5$ s. Peak value $=0.5e^{-1}=0.5(0.3679)=0.1839$. Finite, then decays → confirms stability. ✓

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Ex 6 — Cell F: a zero enters (does it change stability?)

Forecast: Same poles in both. The mistake-callout in the parent warned us: zeros never decide stability. Guess: still stable, only the shape shifts.

1. List poles for both. Why? Poles are the only stability arbiters. Both have poles at $s=-1,-4$ — identical, both left-half → both stable.
2. The zero of $G_2$. Why note it? $s+2=0\Rightarrow$ zero at $s=-2$. It sits between the poles.
3. What the zero does. Partial fractions of $G_2$: the residues (coefficients $A_k$) change compared to $G_1$, so the weights on $e^{-t}$ and $e^{-4t}$ change — the transient shape differs (possible overshoot), but the exponents (and hence stability) are untouched.

Verify: DC gains differ, proving the zero reshapes the response: $G_1(0)=\frac{1}{1\cdot 4}=0.25$; $G_2(0)=\frac{2}{1\cdot 4}=0.5$. Different steady values, same poles. ✓

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Ex 7 — Cell G: a right-half-plane zero (non-minimum-phase)

Forecast: Stable (poles left), but the zero at $s=+2$ is in the right half-plane — the classic "aircraft sinks before it climbs" signature.

1. Poles. Why? Stability check. $(s+1)(s+3)$ → $s=-1,-3$, both left → stable. Good.
2. Zero. Why? $-(s-2)=0\Rightarrow s=+2$, a right-half-plane zero (RHP zero).
3. The wrong-way move. Why does this happen? The initial slope of the step response has the sign of the high-frequency behaviour, while the final value has the sign of the DC gain — and an RHP zero makes these two opposite signs. Check the DC gain: $G(0)=\frac{2}{3}=+0.667$ (settles positive). But the leading coefficient of the numerator is $-s$, so the initial response kicks negative — the vehicle first dips, then rises.

Verify: Initial value theorem on the step response $Y(s)=G(s)\cdot \frac{1}{s}$: initial velocity ${\lim }_{s\to \infty }s\cdot sY(s)={\lim }_{s\to \infty }sG(s)={\lim }_{s\to \infty }\frac{-s^2+2s}{s^2+4s+3}=-1<0$ while $G(0)=+0.667>0$. Opposite signs → wrong-way motion confirmed. ✓

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Ex 8 — Cell H: real-world rocket actuator

Forecast: Middle coefficient positive, constant positive → stable. It looks 2nd-order → probably ringing (Cell B in disguise). Command $1$ rad → should settle near $1$ rad.

1. Laplace with zero I.C. Why zero I.C.? A transfer function describes the actuator itself, so we start from rest. $s^2\Theta +6s\Theta +25\Theta =25U\Rightarrow (s^2+6s+25)\Theta =25U$
2. Form the ratio. Why? $G=\Theta /U$. $G(s)=\frac{25}{s^2+6s+25}$
3. Read $\zeta ,{\omega }_n$. Why? Standard-form matching. ${\omega }_n=\sqrt{25}=5$ rad/s; $2\zeta {\omega }_n=6\Rightarrow \zeta =0.6$. Underdamped → small overshoot, then settles.
4. Poles. $s=-\zeta {\omega }_n\pm j{\omega }_n\sqrt{1-{\zeta }^2}=-3\pm j5(0.8)=-3\pm j4$. Left-half → stable.
5. Steady-state accuracy. Why $s\to 0$? DC gain = final nozzle angle per unit command. $G(0)=25/25=1$ → the nozzle exactly tracks the command in steady state (zero steady-state error).

Verify: $|\text{pole}|=\sqrt{3^2+4^2}=5={\omega }_n$ ✓; $\zeta =3/5=0.6$ ✓; $G(0)=1$ (perfect tracking) ✓.

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Ex 9 — Cell I: limiting values (DC gain and roll-off)

Forecast: At $s\to 0$ we get a finite number (DC gain). At $s\to \infty$ the highest powers win: numerator $\sim 10s$, denominator $\sim s^2$, ratio $\sim 10/s\to 0$ — it rolls off.

1. DC gain, $s\to 0$. Why? Sets the steady response to a constant input. $G(0)=\frac{10(0+1)}{0+0+2}=\frac{10}{2}=5$
2. High-frequency limit, $s\to \infty$. Why? Tells how fast the system attenuates fast signals (roll-off). $G(s)\approx \frac{10s}{s^2}=\frac{10}{s}\to 0$ The system has one more pole (2) than zero (1), so it rolls off like $1/s$ — a $-20$ dB/decade slope on a Bode plot.
3. Interpret. Slow commands pass with gain $5$; fast jitter is filtered out.

Verify: $G(0)=5$ ✓. Excess of poles over zeros $=2-1=1$, so $sG(s)\to 10$ as $s\to \infty$ (finite), confirming a single-order roll-off. ✓

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Ex 10 — Cell J: exam twist, parameter hides instability

Forecast: Increasing $K$ shrinks the constant term $2-K$. Once $2-K$ turns negative, the sign test fails. Guess: stable for $K<2$.

1. Write the denominator coefficients. Why? Use the 2nd-order sign rule: $s^2+a_{1}s+a_0$ has both roots left-half iff $a_1>0$ and $a_0>0$. Here $a_1=3>0$ (always fine) and $a_0=2-K$.
2. Apply the condition. Why? Stability needs $a_0>0$. $2-K>0\Rightarrow K<2$
3. The threshold. At $K=2$: $a_0=0$, so denominator $=s^2+3s=s(s+3)$ → a pole at $s=0$ (on the axis) → marginally stable (Cell D). For $K>2$: $a_0<0$ → a root crosses into the right half → unstable (Cell C).

Verify: At the boundary $K=2$: poles $s=0,-3$; one pole exactly on the imaginary axis → marginal, matching the derivation. At $K=3$: $a_0=-1<0$, roots of $s^2+3s-1$ are $\frac{-3\pm \sqrt{13}}{2}=0.303,-3.303$ → one RHP pole → unstable. ✓

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Recall Which cell was hardest — self-test

An RHP zero makes a system unstable. ::: False — only poles decide stability; an RHP zero (Ex 7) causes wrong-way motion but the system stays stable if poles are left-half. A double pole at $s=-2$ is unstable because of the extra $t$ factor. ::: False — $t{e}^{-2t}\to 0$; still stable (Ex 5). An integrator $1/s$ is stable. ::: No — it is marginally stable; a step input yields an unbounded ramp (Ex 4). For $\frac{K}{s^2+3s+(2-K)}$, the stability limit is ::: $K<2$; at $K=2$ marginal, above it unstable (Ex 10).

Connections

• Transfer function — Laplace domain, poles and zeros — the parent this deep-dive expands
• Second-order systems - damping ratio and natural frequency — used in Ex 2 and Ex 8
• Stability - Routh-Hurwitz criterion — the sign test used in Ex 3 and Ex 10
• Root locus — how poles move as a gain $K$ varies (Ex 10 is one point on such a locus)
• Bode plot — the roll-off computed in Ex 9